We are given two lines which intersect at CC, the line containing the altitude from CC, and the line containing the median from CC. Additionally, we are given the equation of the line containing the median, meaning all that remains is to find the equation of the line containing the altitude, and find their point of intersection.
An altitude of a triangle is a line segment which intersects a vertex of the triangle and the side opposite that vertex, intersecting the side at a right angle.
As the altitude from CC to bar(AB)¯¯¯¯¯¯AB is perpendicular to bar(AB)¯¯¯¯¯¯AB, that means its slope will be equal to the negative reciprocal of the slope of bar(AB)¯¯¯¯¯¯AB. That is, if m_bar(AB)m¯¯¯¯¯¯AB is the slope of bar(AB)¯¯¯¯¯¯AB, then the slope of the altitude will be m_"altitude" = -1/m_bar(AB)maltitude=−1m¯¯¯¯¯¯AB.
The slope of a line containing two points (x_1, y_1)(x1,y1) and (x_2, y_2)(x2,y2) is given by m = (y_2-y_1)/(x_2-x_1)m=y2−y1x2−x1. With that, we can calculate the slope of bar(AB)¯¯¯¯¯¯AB as
m_bar(AB) = (-6-2)/(2-(-5)) = -8/7m¯¯¯¯¯¯AB=−6−22−(−5)=−87
thus
m_"altitude" = -1/m_bar(AB) = 7/8maltitude=−1m¯¯¯¯¯¯AB=78
We are also given that the yy-intercept of the altitude is (0,-5)(0,−5). Thus, using the slope-intercept form of a line, we get the equation of the line containing the altitude as
y = 7/8x - 5y=78x−5
As this intersects the line y = 26/15x + 3/5y=2615x+35 containing the median from CC only at CC, we are looking for the solution to the system of equations
{(y = 7/8x - 5), (y = 26/15x + 3/5):}
=> 7/8x - 5 = 26/15x + 3/5
=> -103/120x = 28/5
=> x = -672/103
=> y = -1103/103
This gives us the final result C(-672/103, -1103/103)
The graph would look as follows, with green dashed line containing the median and the blue dashed line containing the altitude.
![enter image source here]()
